There are 46 students in a math class, there is a class, and the number of students in the class is

24 days, Saturday (if Sundays and holidays are not counted, it is said that they can not be considered) 46 8 = 5 ......Remainder 6

In other words, the top 2 of these 8 students are included in the students on duty after 6 days. In this way, every 6 days will include the next 2 students.

That is, after 4 6 days, these 8 students can be on duty together.

About three weeks and 21 days, launched for Saturday.

46 classmates, 8 people at a time, five days later (the sixth day) two of these 8 people do it with numbers 41-46. In this way, each "reincarnation" of 2 people passed, and in the fourth "reincarnation" there were these eight people again. A total of 24 days passed, so it was Tuesday.

Or code 46 classmates, 46 can be divisible by 8, and the number of reincarnations is 4 times, so 46 * 4 = 184, these 46 classmates are regarded as 184 samples, and every eight people are regarded as a group. 184 samples 8 students in one group = 24 groups. In other words, when it came to the "twenty-fourth group", it was the same eight classmates again.

24 days, including 4 "six days", so it's Wednesday minus one day, it's Tuesday.

The list is as follows: Wednesday 1 8

IV 9 16

V 17-24

VI 25 32

A 33 40

II 40 46 2

III 3 10

IV 11-18

V. 19-26

Liu 27-34

A 35-42

And 43 46 4

iii. 5, 12, S, 13, 20

V. 21-28

Liu 29-36

A 37-44

And 45 46 6

iii. 7, 14, S, 15, 22

V. 23-30

Liu 31-38

A 39, 46, and 18

Original question: There are 55 students in a class, 40 people are good at stools, 42 people are good at Chinese, 44 people are good at foreign languages, and 21 people are good at these three subjects, so how many people are not good at all three?

Answer: First of all, simplify the question, because there are 21 people who are good at all three subjects, so the question is equivalent to:

There are a total of 55-21 = 34 students in a certain class, of which 40-21 = 19 students are good at mathematics, 42 21 21 students are good at Chinese, and 44 21 23 students are good at foreign language.

Therefore, each person may have a maximum of 2 good subjects, and the total number of possible "good" is mostly 34 2 = 68 > 19 + 21 + 23 = 63, so there may be people who are not good at all three, and the maximum is (68-63) 2 2 (rounded).

For example: good at both math and Chinese: 8 students.

Good at math and English at the same time: 11 students.

Good at both Chinese and English: 12 people.

1 person who is good at language.

2 people who are not good at all three.

Good at both math and Chinese: 9 students.

Good at math and English at the same time: 10 students.

Good language and English at the same time: 11 people.

Good English for 2 people.

1 person who is good at language.

Sanmen Xinbi is not good 1 person.

Hopefully, thank you.

There are 40 students in this class.

The denominator is 5 and 4, find the least common multiple of 5 and 4.

4x5=20 (person).

Because the number of students is 30 45 people, so 20x2 = 40 (people) Answer: There are 40 students in this class.

Analysis] This question examines the application of least common multiples.

It can be seen from the question that there are 30 45 students in a class, and on the weekend, 2 5 of the students in the class go to play ball, and 1 4 go to sing, indicating that the number of students in the class is a multiple of 5 and a multiple of 4, and the least common multiple of 4 and 5 is 20, because the number of students in the class is more than 30, less than 45, 20x2 = 40 (people), so there are 40 students in this class.

Those who only answered the first question correctly:

36-27=9 (person).

Those who only answered the second question correctly were:

48-27=21 (person).

Both questions are empty bucket wheels, and the answers are incorrect

60-9-21-27 = 3 (people).

There are three people who are not right.

The number of people who can do the first question plus the number of people who can do the second question and subtract the class size to get the overlapping number of people, that is, the number of people who can do both the first question and the second question.

34 + 27-56 = 5 (people).

Answer: There are 5 people who got both questions right.

Niu, it's okay if you don't finish the fight, I'll go on, if there are x people who have excelled in the two exams, how much is x?

1: If there are x students who are excellent, and Y people who are not excellent, the number of students in the class is (x+y), which is derived from the title:

95x+80y＞90(x+y)

To simplify: x 2y

x/(x+y)＞2y/3y=2/3

That is, the proportion of students who are the best in the class is at least 2 3

2: (21+26+17)-50=14 people.

There were 50 students in the May 1st class, 21 of whom were excellent in the first exam, 26 in the second exam, and 17 who did not excel in both exams. If there are x people in both exams, what is x? That's it.

The number of people who won both times is x, so 50-(21+26-x)=17 x=14 (21+26-x) is the total number of people who won both times but did not repeat, so the number of people who won both times is 14.

The question doesn't seem to be complete, oh, what is your problem?

If there are 2 students, the class size will be a common multiple of , with a minimum of 210

210-2=208 (person).

Hello mchardonnay :

5 people in a row less: 5 3 2 (people).

6 people in a row less: 6 4 2 (people).

7 people in a row less: 7 5 2 (people).

This class has: 5 6 7 2 208 (people).

Math problem: There are more than 2 students in a group of 6 students in a class, and more than 4 students in a group of 8 people, how many people are in this class? The solution is as follows:

Set up a class divided into X groups, and add 2 people to a group of 6 people, and subtract 4 people to a group of 8 people;

6x+2=8x-4

8x-6x=6

2x=6x=36 3+2=20 (person).

20÷8=2...4 (person).

A: There are 20 students in this class;